An object has a mass of #2 kg#. The object's kinetic energy uniformly changes from #32 KJ# to #80 KJ# over #t in [0, 4 s]#. What is the average speed of the object?

Question

1645 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-08-16T08:38:04

The average speed is #=234.8ms^-1#

The kinetic energy is

#KE=1/2mv^2#

The mass is #=2kg#

The initial velocity is #=u_1ms^-1#

The final velocity is #=u_2 ms^-1#

The initial kinetic energy is #1/2m u_1^2=32000J#

The final kinetic energy is #1/2m u_2^2=80000J#

Therefore,

#u_1^2=2/2*32000=32000m^2s^-2#

and,

#u_2^2=2/2*80000=80000m^2s^-2#

The graph of #v^2=f(t)# is a straight line

The points are #(0,32000)# and #(4,80000)#

The equation of the line is

#v^2-32000=(80000-32000)/4t#

#v^2=12000t+32000#

So,

#v=sqrt((12000t+32000)#

We need to calculate the average value of #v# over #t in [0,4]#

#(4-0)bar v=int_0^4(sqrt(12000t+32000))dt#

#4 barv=[((12000t+32000)^(3/2)/(3/2*12000))]_0^4#

#=((12000*4+32000)^(3/2)/(18000))-((12000*0+32000)^(3/2)/(18000))#

#=80000^(3/2)/18000-32000^(3/2)/18000#

#=939.1#

So,

#barv=939.1/4=234.8ms^-1#

The average speed is #=234.8ms^-1#